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The rate of a reaction triples when temp...

The rate of a reaction triples when temperature change from `20^@C " to " 50^@C` . Calculate the energy of activation.

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The correct Answer is:
28.81
`log k_(2)/k_(1) = E_(a)/(R xx 2.303) [(T_(2)-T_(1))/(T_(1)T_(2))]`
Given, `k_(2)/k_(1) =3, R= 8.314 JK^(-1) "mol"^(-1)`
`T_(1) = 20 + 273 = 293 K` and `T_(2) = 50 + 273 = 323 K`
Substituting the given values in the Arrhenium equation,
`log 3 = E_(a)/(8.314 xx 2.303) [(323 - 293)/(323 xx 293)]`
`E_(a) =(2.303 xx 8.314 xx 323 xx 293 xx 0.477)/30`
`=28.81 kJ mol^(-1)`
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